Question

NO LINKS!!! Exponential Growth and Decay Part 4​

Answer 1

#9

`\\ \rm\Rrightarrow P=P_o(1+r/n)^t`

we need t

`\\ \rm\Rrightarrow P=500(1+0.034/52)^t`

`\\ \rm\Rrightarrow 700=500(1+0.000654)^t`

`\\ \rm\Rrightarrow 700=500(1.000654)^t`

`\\ \rm\Rrightarrow (7)/(5)=(1.000654)^t`

`\\ \rm\Rrightarrow 1.4=1.000654^t`

`\\ \rm\Rrightarrow log1.4=log1.000654^t`

`\\ \rm\Rrightarrow log1.4=tlog1.000654`

`\\ \rm\Rrightarrow t=(log1.4)/(1.000654)`

`\\ \rm\Rrightarrow t\approx 514.6\approx 515weeks`

#2

`\\ \rm\Rrightarrow P=240000(1+0.012/25)^t`

`\\ \rm\Rrightarrow 300000=240000(1.0005)^t`

`\\ \rm\Rrightarrow (5)/(4)=(1.0005)^t`

`\\ \rm\Rrightarrow 1.25=(1.0005)^t`

`\\ \rm\Rrightarrow log1.25=log1.0005^t`

`\\ \rm\Rrightarrow log1.25=tlog1.0005`

`\\ \rm\Rrightarrow t=(log1.25)/(1.0005)`

`\\ \rm\Rrightarrow t=446.4\approx 446`

• t=223months

Question is bit unclear to assume final units of time

Answer 2

Problem 9

a = 500 = initial amount

b = 1+r/n = 1 + 0.034/52 = 1.000654 = approximate growth factor

The equation goes from
`y = a*b^x` to
`y = 500*1.000654^x` where x is the number of weeks and y is the account balance in dollars.

Plug in y = 700 and isolate x by using logarithms.

`y = a*b^x\\\\y = 500*1.000654^x\\\\700 = 500*1.000654^x\\\\700/500 = 1.000654^x\\\\1.4 = 1.000654^x\\\\1.000654^x = 1.4\\\\\log(1.000654^x) = \log(1.4)\\\\x\log(1.000654) = \log(1.4)\\\\x = \log(1.4)/\log(1.000654)\\\\x = 514.652\\\\x = 515\\\\`

It takes about 515 weeks for the account balance to reach $700

This is equivalent to roughly 515/52 = 9.904 years

Answer: 515 weeks

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Problem 10

The formula to calculate the monthly payment is not a simple exponential growth or decay function. I'm not sure how your teacher wanted you to answer this question. You'll have to get clarification on it.

If your teacher is asking something along the lines of "$240,000 is invested at a rate of 1.2% compounded bimonthly. How many years does it take to have the investment reach $300,000?" then you could follow these steps below.

a = 240,000 = initial amount

b = 1 + r/n = 1 + 0.012/24 = 1.0005 = growth factor

`y = a*b^x\\\\y = 240000*1.0005^x\\\\300000 = 240000*1.0005^x\\\\240000*1.0005^x = 300000\\\\1.0005^x = 300000/240000\\\\`

`1.0005^x = 1.25\\\\\log(1.0005^x) = \log(1.25)\\\\x\log(1.0005) = \log(1.25)\\\\x = \log(1.25)/\log(1.0005)\\\\x = 446.3987\\\\x = 446\\\\`

The x represents the number of bimonthly periods. There are 24 bimonthly periods in a year, which means it takes roughly 446/24 = 18.58 years for the starting balance of $240,000 to reach $300,000.

Keep in mind that the steps above are based on the assumption I made at the top of the problem. If that assumption is incorrect, then ignore the steps and answer entirely.